To evaluate the integral using the indicated trigonometric substitution, we'll proceed as follows:

Step 1: Identify the Substitution

Given $x = 3 \tan(\theta)$, we need to calculate $dx$ and substitute into the integral.

• Substitute $x = 3 \tan(\theta)$: $dx = 3 \sec^2(\theta) \, d\theta$

• Simplify the expression under the square root: $x^2 + 9 = (3 \tan(\theta))^2 + 9 = 9 \tan^2(\theta) + 9 = 9 (\tan^2(\theta) + 1) = 9 \sec^2(\theta)$ So, $\sqrt{x^2 + 9} = 3 \sec(\theta)$.

Step 2: Substitute into the Integral

The original integral is: $\int \frac{8x^3}{\sqrt{x^2 + 9}} \, dx$ Substitute $x = 3 \tan(\theta)$ and $dx = 3 \sec^2(\theta) \, d\theta$: $\int \frac{8(3 \tan(\theta))^3}{3 \sec(\theta)} \cdot 3 \sec^2(\theta) \, d\theta$ Simplifying: $\int \frac{8 \cdot 27 \tan^3(\theta)}{3} \cdot \sec(\theta) \, d\theta = 72 \int \tan^3(\theta) \sec(\theta) \, d\theta$

Step 3: Simplify the Integral

Recall that $\tan^3(\theta) = \tan(\theta) \cdot \tan^2(\theta)$ and $\tan^2(\theta) = \sec^2(\theta) - 1$: $72 \int \tan(\theta)(\sec^2(\theta) - 1) \sec(\theta) \, d\theta$ Expanding: $72 \int (\tan(\theta) \sec^3(\theta) - \tan(\theta) \sec(\theta)) \, d\theta$ This can be split into two integrals: $72 \left( \int \tan(\theta) \sec^3(\theta) \, d\theta - \int \tan(\theta) \sec(\theta) \, d\theta \right)$

Step 4: Evaluate the Integrals

1. Integral of $\tan(\theta) \sec(\theta)$: $\int \tan(\theta) \sec(\theta) \, d\theta = \frac{\sec^2(\theta)}{2} + C$

2. Integral of $\tan(\theta) \sec^3(\theta)$: This requires a more advanced method of integration, typically using reduction formulas or rewriting in terms of sine and cosine. After evaluating, we find:

$\int \tan(\theta) \sec^3(\theta) \, d\theta = ...$ Continue the solution